Current Parts:
Caps = 22uf
Transistors = 2N3904
Resistors = (2) 1k brown-black-red
Resistors = (2) 15k brown-green-orange
LEDs = (1st side) r, r, w, r, w, b
LEDs = (2nd side) r, r, w, r, w, b
Power = 9 volt batt.

My problem is that as it stands, I can not get more than 2 of the LEDs to alternate as you see them connected in series. All of the LEDs I am using are from a Christmas light set (set of 60 for $5.00 was awesome from BigLots after Xmas clearance sale), none of which have resistors and that is where I am guessing I need the help or within the circuit possibly. I am also open to changing the power source - not to exceed 9 volts though. So if anyone can help me - I'd love it! Thanks. I am a newbie to this obviously.

P.S. - I also have a couple 555 and 4017 IC's on hand if needed.

 

Hello,

Can you give the schematic ?
A schematic tells more than 1000 words.

Greetings,
Bertus

 

Hello,

Can you give the schematic ?
A schematic tells more than 1000 words.

Greetings,
Bertus

It is simliar to this circuit but with my values and needs:

 

Hello,

The 9 Volts will limit the number of leds connected in series.
The different leds have different voltage drops, see the table in the link.
http://www.oksolar.com/led/led_color_chart.htm

Greetings,
Bertus

 

Hello,

The 9 Volts will limit the number of leds connected in series.
The different leds have different voltage drops, see the table in the link.
http://www.oksolar.com/led/led_color_chart.htm

Greetings,
Bertus

I do know that already - I was hoping for a little more detailed help. I think I need a resistor on each LED. It occurs to me that I may only be able to light 3 LEDs per side with 9volt no matter what resistors are used - I wonder if this is correct. Remember, I am a novice. But thank you anyway.

 

Hello,

You can calculate the resistor.
Take the voltage from the power supply, subtract the voltage drop of the leds.
Divide this difference by the desired led current.

Example : powersupply = 9 Volts, led voltage = 2.2 for the red and 3.8 for the blue led in the string.
So the voltage difference is 9 - (2.2 + 3.8) = 9 - 6 = 3 volts.
You want 20 ma as led current. So the resistor is 3 / 0.02 = 150 Ohms.

I do not have the voltage drop of the transistor (1 volt max) in the calculation, so the current will be a little lower.

Greetings,
Bertus

 

Hello,

You can calculate the resistor.
Take the voltage from the power supply, subtract the voltage drop of the leds.
Divide this difference by the desired led current.

Example : powersupply = 9 Volts, led voltage = 2.2 for the red and 3.8 for the blue led in the string.
So the voltage difference is 9 - (2.2 + 3.8) = 9 - 6 = 3 volts.
You want 20 ma as led current. So the resistor is 3 / 0.02 = 150 Ohms.

I do not have the voltage drop of the transistor (1 volt max) in the calculation, so the current will be a little lower.

Greetings,
Bertus

Ok I gather what you are saying. I want to just get two LEDs lit for now. The blue and the white. So 3.8 + 3.6 = 7.4 ... so 9 - 7.4 = 1.6 ... and 1.6 / .02 = 80ohm. I am going to try this.....

 

Hello,

You are correct. A resistor of 80 Ohms does not exist so you take the first larger in a E-series.
Here is a list of standard resistor values:
http://www.elexp.com/t_rescht.htm

This page comes from the component data page from the EDUCYPEDIA.
http://www.educypedia.be/electronics/datacomponent.htm

Greetings,
Bertus

 

Hello,

You are correct. A resistor of 80 Ohms does not exist so you take the first larger in a E-series.
Here is a list of standard resistor values:
http://www.elexp.com/t_rescht.htm

This page comes from the component data page from the EDUCYPEDIA.
http://www.educypedia.be/electronics/datacomponent.htm

Greetings,
Bertus

Yeah 82, 91 and 100ohm are next. I have 100ohms and will use...